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Proof step 5 from proof newrefs-three (previous step)

⊦ newref_list(2, st[newref(st) -- 0 ', noval]) = newref(st[newref(st) -- 0 ', noval]) ' + newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < n → newref_list(n, st) = newref(st) + newref_list(n -1, st[newref(st) -- 0 ', noval])
Proof step 4:
⊦ newref_list(1, st) = newref(st) '

Rule argument

[crewrite-arg: newrefs-rec-01 , F , [st, n] → [st[newref(st) -- 0 ', noval], 2] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r + refs = r ' + refs], [CSIMPSEQ : ⊦ refs + refs0 = refs + refs1 ↔ refs0 = refs1], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1])], [EXTRATERMS : (st[newref(st) -- 0 ', noval], 2)]]